Question

2. The diagonal of a rectangle is 10 cm and its breadth is 6 cm. Find its (1) length (ü) area.​

Answer 1

Answer:

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Answer 2

To Find :-

• The Length of the Rectangle.

• The Area of the Rectangle.

Given :-

• Breadth of the Rectangle = 6 cm.

• Diagonal of the Rectangle = 10 cm.

We Know :-

Diagonal of the Rectangle :-

$\boxed{\underline{\bf{D = \sqrt{l^{2} + b^{2}}}}}$

Where :-

• l = Length of the Rectangle

• b = Breadth of the Rectangle

• D = Diagonal of the Rectangle

Area of a Rectangle :-

$\boxed{\underline{\bf{A = length \times Breadth}}}$

Solution :-

Diagonal of the Rectangle :-

Using the formula and substituting the values in it , we get :-

$:\implies \bf{D = \sqrt{l^{2} + b^{2}}} \\ \\ \\ :\implies \bf{10 = \sqrt{l^{2} + 6^{2}}} \\ \\ \\ :\implies \bf{10^{2} = l^{2} + 6^{2}} \\ \\ \\ :\implies \bf{10^{2} - 6^{2} = l^{2}} \\ \\ \\ :\implies \bf{\sqrt{10^{2} - 6^{2}} = l} \\ \\ \\ :\implies \bf{\sqrt{100 - 36} = l} \\ \\ \\ :\implies \bf{\sqrt{64} = l} \\ \\ \\ :\implies \bf{8 cm = l} \\ \\ \\ \therefore \purple{\bf{l = 8 cm}}$

Hence, the Length of the Rectangle is 8 cm.

Area of the Rectangle :-

Given :-

• Length = 8 cm

• Breadth = 6 cm

Using the formula and substituting the values in it , we get :-

$:\implies \bf{A = length \times Breadth} \\ \\ \\ \implies \bf{A = 8 \times 6} \\ \\ \\ \implies \bf{A = 48 cm^{2}} \\ \\ \\ \therefore \purple{\bf{A = 48 cm^{2}}}$

Hence, the Area of the Rectangle is 48 cm².